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Finitely Generated Abelian Groups

  1. Mar 4, 2012 #1
    1. The problem statement, all variables and given/known data

    Let [tex] p [/tex] be prime and let [tex] b_1 ,...,b_k [/tex] be non-negative integers. Show that if :
    [tex] G \simeq (\mathbb{Z} / p )^{b_1} \oplus ... \oplus (\mathbb{Z} / p^k ) ^ {b_k} [/tex]
    then the integers [tex] b_i [/tex] are uniquely determined by G . (Hint: consider the kernel of the homomorphism [tex] f_i :G \to G [/tex] that is multiplication by [tex] p^i [/tex] . Show that [tex] f_1 , f_2 [/tex] determine [tex] b_1[/tex].
    Proceed similarly )


    2. Relevant equations
    The question, together with the theorem it should help me prove is attached (this question should help me prove the uniqueness part of the theorem)


    3. The attempt at a solution
    I've tried using the hint, and considering the homomorphism [tex] f_1 : G \to G [/tex] that is defined by [tex]f_1 (x ) = px [/tex] . Its kernel is [tex] ker f_1 = (\mathbb{Z}_p ) ^{b_1} \bigoplus (\mathbb{Z}_p) ^ {b_2} ... [/tex].
    But how does it help me? Am I right in my calculation of the kernel?


    Hope someone will be able to help me

    Thanks in advance !
     

    Attached Files:

  2. jcsd
  3. Mar 9, 2012 #2

    morphism

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    Re: Finitely-Generated-Abelian-Groups

    The hint is essentially leading you to consider a proof by induction on k. Notice that $$ \ker f_1 = (\mathbb Z/p)^{b_1} \oplus (\mathbb Z/p)^{b_2} \oplus (\mathbb Z/p^2)^{b_3} \oplus \cdots = (\mathbb Z/p)^{b_1+b_2} \oplus (\mathbb Z/p^2)^{b_3} \oplus \cdots. $$
     
  4. Mar 9, 2012 #3
    Re: Finitely-Generated-Abelian-Groups

    Can you please explain how did you get that this is the kernel?

    Thanks a lot !
     
  5. Mar 9, 2012 #4

    morphism

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    Re: Finitely-Generated-Abelian-Groups

    Well, if px=0 mod p^i, then p^i divides px, hence p^{i-1} divides x, i.e. x=0 mod p^i.
     
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